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Math Help - Proving an identity

  1. #1
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    Proving an identity

    Establish the identity:

    (cotθ / 1- tanθ) + (tanθ / 1 – cotθ) = 1+ tanθ + cotθ
    Last edited by mr fantastic; March 26th 2009 at 07:50 PM. Reason: Moved post and edited
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  2. #2
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    Trig Identity

    Hello xhazel06x
    Quote Originally Posted by xhazel06x View Post
    Establish the identity:

    (cotθ / 1- tanθ) + (tanθ / 1 – cotθ) = 1+ tanθ + cotθ
    If we let \tan\theta = t, then

    \frac{\cot\theta}{1-\tan\theta}+\frac{\tan\theta}{1-\cot\theta}= \frac{\frac{1}{t}}{1-t}+\frac{t}{1-\frac{1}{t}}

    =\frac{\frac{1}{t}}{1-t}+\frac{t^2}{t-1}

    =\frac{\frac{1}{t}}{1-t}-\frac{t^2}{1-t}

    =\frac{\frac{1}{t}-t^2}{1-t}

    =\frac{\frac{1}{t}(1-t^3)}{1-t}

    =\frac{\frac{1}{t}(1-t)(1+t+t^2)}{1-t}

    =\frac{1}{t} + 1 + t

    = 1 + \tan\theta+\cot\theta

    Grandad
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  3. #3
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    Hello, xhazel06x!

    Another approach . . . longer than Grandad's


    Establish the identity: . \frac{\cot\theta}{1-\tan\theta} + \frac{\tan\theta}{1-\cot\theta} \:=\: 1+ \tan\theta + \cot\theta
    The left side is: . \frac{\dfrac{\cos\theta}{\sin\theta}}{1 - \dfrac{\sin\theta}{\cos\theta}} + \frac{\dfrac{\sin\theta}{\cos\theta}} {1 - \dfrac{\cos\theta}{\sin\theta}}


    Multiply the fractions by \frac{\sin\theta\cos\theta}{\sin\theta\cos\theta}

    \frac{\sin\theta\cos\theta\left(\dfrac{\cos\theta}  {\sin\theta}\right)}{\sin\theta\cos\theta\left(1 - \dfrac{\sin\theta}{\cos\theta}\right)} + \frac{\sin\theta\cos\theta\left(\dfrac{\sin\theta}  {\cos\theta}\right)} {\sin\theta\cos\theta\left(1 - \dfrac{\cos\theta}{\sin\theta}\right)} . = \;\frac{\cos^2\!\theta}{\sin\theta\cos\theta - \sin^2\!\theta} + \frac{\sin^2\!\theta}{\sin\theta\cos\theta - \cos^2\!\theta}

    . . =\;\frac{\cos^2\!\theta}{\sin\theta(\cos\theta-\sin\theta)} + \frac{\sin^2\!\theta}{-\cos\theta(\cos\theta - \sin\theta)} . = \;{\color{blue}\frac{\cos\theta}{\cos\theta}}\cdot  \frac{\cos^2\!\theta}{\sin\theta(\cos\theta - \sin\theta)} \;- \;{\color{blue}\frac{\sin\theta}{\sin\theta}}\cdot \frac{\sin^2\!\theta}{\cos\theta(\cos\theta - \sin\theta)}

    . . = \;\frac{\cos^3\!\theta}{\sin\theta\cos\theta(\cos\  theta-\sin\theta)} - \frac{\sin^3\!\theta}{\sin\theta\cos\theta(\cos\th  eta - \sin\theta)} . = \;\frac{\cos^3\!\theta - \sin^3\theta}{\sin\theta\cos\theta(\cos\theta - \sin\theta)}

    . . = \;\frac{(\cos\theta-\sin\theta)(\cos^2\!\theta + \cos\theta\sin\theta + \sin^2\!\theta)}{\sin\theta\cos\theta(\cos\theta - \sin\theta)} . = \;\frac{\cos^2\!\theta + \cos\theta\sin\theta + \sin^2\!\theta}{\sin\theta\cos\theta}

    . . =\; \frac{\cos^2\!\theta}{\sin\theta\cos\theta} + \frac{\cos\theta\sin\theta}{\sin\theta\cos\theta} + \frac{\sin^2\!\theta}{\sin\theta\cos\theta} . =\; \frac{\cos\theta}{\sin\theta} + 1 + \frac{\sin\theta}{\cos\theta}


    . . =\;\cot\theta + 1 + \tan\theta

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